标签:val incr different lines star mis like tianjin hdu
2 5 7 3 3 3 0 3 1 0 0 4 5 1 3 3 2 3 4 2 4 3 1 5 6 4 5 3 1 4 4 3 4 2 6 7 -1 -1 0 1 0 2 1 0 1 1 2 3 1 2 1 2 3 6 4 5 5 5 6 3 1 4 6 2 5 5 3 6 4
9 6
题意:有N个岛屿 M条无向路 每一个路有一最大同意的客流量。求从最西的那个岛屿最多能运用多少乘客到最东的那个岛屿。
题解:最大流,起点为最左的点,终点为最右的点。
#include<algorithm> #include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #define N 100020 #define ll long long using namespace std; const int MAXN = 100010;//点数的最大值 const int MAXM = 400010;//边数的最大值 const int INF = 0x3f3f3f3f; struct Edge { int to,next,cap,flow; } edge[MAXM]; //注意是MAXM int tol; int head[MAXN]; int gap[MAXN],dep[MAXN],cur[MAXN]; int n,m; void init() { tol = 0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int w,int rw = 0) { edge[tol].to = v; edge[tol].cap = w; edge[tol].flow = 0; edge[tol].next = head[u]; head[u] = tol++; edge[tol].to = u; edge[tol].cap = rw; edge[tol].flow = 0; edge[tol].next = head[v]; head[v] = tol++; } int Q[MAXN]; void BFS(int start,int end) { memset(dep,-1,sizeof(dep)); memset(gap,0,sizeof(gap)); gap[0] = 1; int front = 0, rear = 0; dep[end] = 0; Q[rear++] = end; while(front != rear) { int u = Q[front++]; for(int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].to; if(dep[v] != -1)continue; Q[rear++] = v; dep[v] = dep[u] + 1; gap[dep[v]]++; } } } int S[MAXN]; int sap(int start,int end,int n) { BFS(start,end); memcpy(cur,head,sizeof(head)); int top = 0; int u = start; int ans = 0; while(dep[start] < n) { if(u == end) { int Min = INF; int inser; for(int i = 0; i < top; i++) if(Min > edge[S[i]].cap - edge[S[i]].flow) { Min = edge[S[i]].cap - edge[S[i]].flow; inser = i; } for(int i = 0; i < top; i++) { edge[S[i]].flow += Min; edge[S[i]^1].flow -= Min; } ans += Min; top = inser; u = edge[S[top]^1].to; continue; } bool flag = false; int v; for(int i = cur[u]; i != -1; i = edge[i].next) { v = edge[i].to; if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]) { flag = true; cur[u] = i; break; } } if(flag) { S[top++] = cur[u]; u = v; continue; } int Min = N; for(int i = head[u]; i != -1; i = edge[i].next) if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) { Min = dep[edge[i].to]; cur[u] = i; } gap[dep[u]]--; if(!gap[dep[u]])return ans; dep[u] = Min + 1; gap[dep[u]]++; if(u != start)u = edge[S[--top]^1].to; } return ans; } int main() { //freopen("test.in","r",stdin); int t; cin>>t; while(t--) { scanf("%d%d",&n,&m); int s,t; int xmin=INF,xmax=-INF; int x,y,c; for(int i=1; i<=n; i++) { scanf("%d%d",&x,&y); if(x<xmin) { xmin=x; s=i; } if(x>xmax) { xmax=x; t=i; } } init(); for(int i=1; i<=m; i++) { scanf("%d%d%d",&x,&y,&c); addedge(x,y,c,c); } printf("%d\n",sap(s,t,n)); } return 0; }
Hdu 4280 Island Transport(最大流)
标签:val incr different lines star mis like tianjin hdu
原文地址:http://www.cnblogs.com/yxysuanfa/p/6999271.html